= Solution
Take $x$ down the plane and $y$ normally away from it, with the solid at $y=0$ and the <free surface> at $y=h$. A steady unidirectional <falling film flow> has velocity $\mathbf u=u(y)\mathbf e_x$. The streamwise <Navier-Stokes equation> reduces to
$$
\nu u''=-g\sin\alpha,
$$
where $\nu$ is the <kinematic viscosity>. The lower <no-slip boundary condition> and upper <stress-free boundary condition> are
$$
u(0)=0,
\qquad
u'(h)=0.
$$
Integrating gives the velocity profile
$$
\boxed{u(y)=\frac{g\sin\alpha}{\nu}\left(hy-\frac{y^2}{2}\right)}.
$$
The <volume flux> per unit width is
$$
\boxed{Q=\int_0^hu(y)\,dy=\frac{g h^3\sin\alpha}{3\nu}}.
$$
If the upper surface is replaced by a stationary solid plane, the second boundary condition becomes $u(h)=0$. The resulting plane-channel profile and flux are
$$
u(y)=\frac{g\sin\alpha}{2\nu}y(h-y),
\qquad
\boxed{Q_{\rm closed}=\frac{g h^3\sin\alpha}{12\nu}=\frac14Q}.
$$
The free surface exerts no tangential <shear stress>, whereas the stationary upper wall enforces no slip and exerts a retarding shear. This additional drag accounts for the reduced flux.
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